Chem Differential Eq HW Solutions Fall 2011 80

Chem Differential Eq HW Solutions Fall 2011 80 - 80 Chapter...

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Unformatted text preview: 80 Chapter 5 Partial Differential Equations in Spherical Coordinates Solutions to Exercises 5.1 1. Start with Laplace’s equation in spherical coordinates (1) ∂u ∂2u ∂ 2 u 2 ∂u 1 ∂2u + + cot θ +2 + csc2 θ ∂r2 r ∂r r ∂θ2 ∂θ ∂φ2 = 0, where 0 < r < a, 0 < φ < 2π, and 0 < θ < π. To separate variable, take a product solution of the form u(r, θ, φ) = R(r)Θ(θ)Φ(φ) = RΘΦ, and plug it into (1). We get 2 1 R ΘΦ + R ΘΦ + 2 RΘ Φ + cot θRΘ Φ + csc2 θ RΘΦ r r = 0. Divide by RΘΦ and multiply by r2 : r2 R Φ R Θ Θ + 2r + + cot θ + csc2 θ = 0. R R Θ Θ Φ Now proceed to separate the variables: r2 R R + 2r =− R R Φ Θ Θ + cot θ + csc2 θ Θ Θ Φ . Since the left side is a function of r and the right side is a function of φ and θ, each side must be constant and the constants must be equal. So r2 R R + 2r =µ R R and Θ Φ Θ + cot θ + csc2 θ = −µ. Θ Θ Φ The equation in R is equivalent to (3). Write the second equation in the form Θ Φ Θ + cot θ + µ = − csc 2 θ ; Θ Θ Φ sin2 θ Θ Θ + cot θ +µ Θ Θ =− Φ . Φ This separates the variables θ and φ, so each side must be constant and the constant must be equal. Hence sin2 θ Θ Θ + cot θ +µ Θ Θ =ν and Φ ⇒ Φ + ν Φ = 0. Φ We expect 2π-periodic solutions in Φ, because φ is and azimuthal angle. The only way for the last equation to have 2π-periodic solutions that are essentially different is to set ν = m2 , where m = 0, 1, 2, . . .. This gives the two equations ν=− Φ + m2 Φ = 0 (equation (5)) and sin2 θ which is equivalent to (6). Θ Θ + cot θ +µ Θ Θ = m2 , ...
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